The Klein quatic is tiled by 24 regular heptagons with interior angles \(2\pi / 3\) . Take any geodesic segment \(\sigma \)  that joins the midpoints of two sides that are separated by one other side in such a heptagon. By extending \(\sigma \)  in both directions, we obtain a primitive closed geodesic that crosses a total of 6 heptagons in the same way (joining the midpoints of two sides separated by another). One example of such a geodesic is depicted in Figure REF . The fact that this particular example closes up implies that all others do since the isometry group of the Klein quartic acts transitively on flags of heptagons in the tiling, hence on geodesic segments with the same description as \(\sigma \) . There are 7 geodesic segments to choose from per heptagon and 24 heptagons, while each closed geodesic obtained consists of 6 segments, for a total of \(7\times 24 / 6 = 28\)  closed geodesics of this kind.
<FIGURE>The length of these closed geodesics is equal to 6 times the length of \(\sigma \) . Note that \(\sigma \)  forms an isoceles triangle with the center of its heptagon \(H\) , with two sides of length equal to the inner radius \(r\)  of \(H\)  (the second shortest side of a \((2,3,7)\) -triangle), and angle \(4\pi /7\)  at the center of \(H\) . Trigonometric formulas for right-angled triangles [1]} yield
\(r = \operatorname{arccosh}(\cos (\pi /3)/\sin (\pi / 7)) = \operatorname{arccosh}(1/(2\sin (\pi / 7)))\) 
and
\(\ell (\sigma )/2 = \operatorname{arcsinh}(\sin (2\pi /7) \sinh (r)).\) 
Since \(\sinh (\operatorname{arccosh}(y)) = \sqrt{y^2 - 1}\) , we have \(\sinh (r)=\sqrt{\frac{1}{4\sin ^2(\pi /7)}-1}\) . Combining this with \(\sin (2\pi / 7)= 2 \sin (\pi /7)\cos (\pi /7)\)  yields the folmula
\(L = 6\, \ell (\sigma ) = 12 \operatorname{arcsinh}\left(\cos (\pi /7)\sqrt{1-4\sin ^2(\pi /7)}\right)\) 
for the length \(L\)  of the resulting geodesics. Using the identity \(\cosh (2x)=2\sinh ^2(x)+1\) , we get that \(2\operatorname{arcsinh}(y) = \operatorname{arccosh}(2y^2+1)\)  for \(y>0\) , and hence
\(L &= 6 \operatorname{arccosh}\left(2\cos ^2(\pi /7)(1-4\sin ^2(\pi /7))+1\right)\\& = 6 \operatorname{arccosh}\left(2\cos ^2(\pi /7)(4\cos ^2(\pi /7)-3)+1\right)\\&= 6 \operatorname{arccosh}\left(8\cos ^4(\pi /7)-6\cos ^2(\pi /7)+1\right).\) 
